[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 6 from proof eq (previous step)

m1 * m0 + m0 = n * m0, ∀ m0, n. m1 * m0 = n * m0 → m0 = 0 ∨ m1 = n, m1 * m0 = (n - 1) * m0 → m1 = n - 1 ⊦ m0 = 0, m1 + 1 = n

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m